Globe Impossible! Coriolis Force and Airplanes

Does the Coriolis Force on airplanes flying North prove the Globe is impossible?

This is almost a shaggy dog story so I'll spoil the ending here for the tl;dr -- it acts like something between gentle breeze to a stiff wind, is corrected the same way, and there is pretty much no net effect as it is constantly countered by holding the heading steady.

What is the Coriolis Force?

Very simply stated, it is an apparent force (not a Proper Force that we measure as an acceleration on a mass) caused by inertia in a rotating reference frame.

The simple example is you are on a merry-go-round and throw a ball towards the center, it appears to be deflected. But from the point of view of someone NOT on the merry-go-round the ball goes in a straight-line once you release it.  So that's why it's not a Proper Force.

Translating that to a globe, if you are on the Equator you have an Eastward velocity of about 464.9 m/s (or 1040 mph).  As you move North in an airplane you retain this initial Eastward velocity but the Earth rotational speed here is slightly less - so you appear to be moving East slightly faster than the ground at this point.

Simple right?

There are actually several ways that this manifests on a rotating sphere like the Earth -- the main effect is the horizontal component resulting from horizontal motion - but there is also the Eötvös effect which causes an object moving Westward to be deflected towards Earth center (making it heavier on a scale) and an objecting moving Eastward would be deflected upwards (making it lighter on a scale).

Wolfie6020 has produced several videos showing this effect as well as the centrifugal change by latitude:



We're just going to look at the horizontal dimension here - a plane flying North and the apparent Eastward deflection.

The whole equation is a vector multiplication where

Coriolis acceleration = -2 Ω × 𝑣

The sign here really doesn't matter -- it just tells you which way on the axis, we just want magnitude anyway.  But let's strip it down to just the north vector (so our east/west and vertical motion is zero) so we end up with:

acceleration = 2 ω sin(φ) 𝑣ₙ

Omega (ω) is simply our rotational rate which is the angle of rotation divided by the time - I'm going to ignore the altitude component of the airplane and just use 1 sidereal day (86164.09054 seconds) over the full circle so that is 2π/86164.09054s.

For our velocity north we will use 𝑣ₙ = 250 m/s (~559.2 mph).

The final component here is φ which is our latitude, the sin(0°) = 0 and sin(90°) = 1 so this simply scales the whole thing by latitude.  Since the maximum effect is at 90° let's look at that value:

2×(250m/s)×(2π/86164.09054s)×sin(90°) ~ 0.03646 m/s² [wolfram|alpha]

Gravity is 9.8 m/s² so this is 1/269th the effect of gravity (and we will show below, equal to no more than a stiff crosswind and far below typical winds aloft).

Let's assume a loaded 439,985 kg 747

That means the maximum force from the above Coriolis acceleration is about 16 kN [wolfram|alpha] from Force = Mass × Acceleration.  For comparison, the force of gravity is 4312.7 kN.

Let's also consider how fast the ground is moving given our latitude:

Tangential speed = distance / time = 2π × R × cos(φ) / 86164.09054s

φ = 0° = 2π × 6378137m × cos(0°) / 86164.09054s ~ 465.1 m/s ~ 1040 mph [wolfram|alpha]

φ = 1° = 2π × 6378137m × cos(1°) / 86164.09054s ~ 465.0 m/s ~ 1040 mph [wolfram|alpha]

φ = 45° = 2π × 6378137m × cos(45°) / 86164.09054s ~ 328.9 m/s ~ 736 mph [wolfram|alpha]

φ = 89° = 2π × 6378137m × cos(89°) / 86164.09054s ~ 8.1 m/s ~ 18 mph [wolfram|alpha]

So our first 1° North is ~69.1 miles and we've only needed to reduce our Eastward velocity by 0.1 m/s (0.22 mph) over about 7.5 minutes of flight time.  This is down in the noise and doesn't even register as a gentle breeze.  The real winds, thermal, and pressure differences will overwhelm any such noise.

The maximum effect is when we're at the Pole -- consider we're about 1° or about 69.1 miles from the North Pole the Earth (and our plane) are both moving East now at just 8.1 m/s (18 mph) and we've flown 6150 miles slowly and continuously reducing our Eastward velocity down.  By the time we fly over the North Pole we reduce it to Zero, and then as we proceed to fly due South from the North Pole we need to (conversely) slowly start increasing our Westward speed to remain on course.  And since we're already crabbed slightly Westward we just begin the slow and continuous reduction of the crab angle.

Now let's consider the effect of a crosswind on the plane.  The specific values will vary of course but we're just look at ONE example in order to help us think about the problem in more general terms.

Air density at 10km (~35,000 feet) (p) is 0.4135kg/m³ [via Engineers Toolbox]

I estimated the profile area of the 747-400 by pixel count at 553 m²

Coefficient of drag I placed between a circular tube and a square tube at 0.6

Wind Speed of 15.3 m/s (about 34 mph, or 30 knots - actual winds aloft can hit over 200 knots)

So we can plug these into our equation for

force of air resistance = ½×p×C×A×v²

and we find the force is about 16 kN

0.5×(0.4135kg/m³)×(0.6)×(553m²)×(15.3m/s)² = 16 kN [wolfram|alpha]

By comparison, a 120 knot crosswind (not uncommon at 35,000 feet) would be on the order of 261 kN of force [wolfram|alpha]

So, the maximum Coriolis force on the plane is about like a stiff wind but far less than typical winds aloft.

In fact, we can treat this exactly like a shift in the wind angle taking us off course by just subtracting the Coriolis vector from the Wind vector -- and indeed the plane cannot tell any difference between the two.  Pilots and autopilots know how to correct for this drift.  If we're drifting Eastward then we turn the plane a bit Westward until our true heading is corrected.  This turning of the plane Westward to counteract this drift would, in turn, automatically cancel out our Eastward excess velocity as we fly.  In addition to that, you are constantly fighting air resistence to keep the plane moving forward -- any velocity changes show up as airspeed changes which are going to cancel out very quickly by just keeping the plane on the correct heading and at the desired altitude and airspeed.

So the cumulative effect is pretty much nil at the end and the instantaneous effect just feels like wind and is easily corrected for.

For East/West motions the deflection is up or down and we handle this by keeping actual vertical speed near zero -- this automatically cancels out the Eötvös effect.  Again, this is a tiny fraction of the gravitational force.

So no issue with Coriolis forces are found for flying airplanes.